Example of a semigroup such that it contains a sub-semigroup which is a non-trivial group

Question

Give an example of a semi-group such that it contains a sub-semigroup which is a non-trivial group.

I am not very sure what a sub-semigroup is but according to Wikipedia:

The semigroup operation induces an operation on the collection of its subsets: given subsets $A$ and $B$ of a semigroup $S$, their product $A·B$, written commonly as $AB$, is the set $\{ ab | a \in A \text{ and } b \in B \}$. (This notion is defined identically as it is for groups.)

In terms of this operation, a subset $A$ is called a subsemigroup if $AA$ is a subset of $A$.

That is they mean $A$ will be a subsemigroup iff $\{a^2|a\in A\}$ is a subset of the set $\{a|a\in A\}$.

I was thinking that since $(\Bbb Z^{+},+)$ is a semigroup, it's subsemigroup $\{1^2,2^2,3^2,...\}$ could be a valid semi-semigroup could be a valid answer. But the problem is that it does not satisfy "closure". Say $2^2+5^2$ isn't an element of $\{1^2,2^2,3^2,...\}$. Any other possible answers to the question?

Answer 1

As a quick recap to be on the same page:

1. A subsemigroup of a semigroup $G$ is a subset $S$ which is closed under the multiplication induced from $G$.

This implies that your subsemigroup $(S,\cdot_{|G})$ is also a semigroup in its own right.

2. A subgroup of a semigroup $G$ is a subsemigroup which happens to be a group.

As for your question, you can construct many example by hand.

But if you want an example quite natural, but not trivial, I'd suggest going for $(\mathcal{M}_2(\mathbb{R}), \cdot)$ the 2x2-matrix semigroup (over $\mathbb{R}$) with matrix multiplication.

Aside from the trivial subgroup $\{\mathbf{0}\}$, you should be able to find another one which is quite famous (hint: it contains the identity matrix $\mathbf{1}$).

Furthermore, interestingly enough, $\mathcal{M}_2(\mathbb{R})$ contains also an infinite family of subgroups which aren't trivial (and they aren't that hard to describe as a set once you've found them).

Answer 2

Any nontrivial group is a semigroup and hence a solution to your question. If you want a group $G$ which is different from the original semigroup, take $S = G \cup \{0\}$, where $0$ is a new element and define the product by $0$ by setting, for all $s \in S$, $s0= 0s = 0$.